3(2 - x) ≥ 2(1 - x) – Class 11 Mathematics

Exercise 5.1 · Q8

3(2 - x) \ge 2(1 - x)

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Concept understanding — Linear Inequality Solutions

Linear Inequality Solutions – A First Look

Imagine you're standing on a number line. You know exactly where the number 5 is. But what if I asked you to stand on "all numbers greater than 5"? You can't stand on all of them at once — they stretch infinitely to the right. That's the core idea of an inequality: instead of one exact point, you get a whole region of possible values.

A linear inequality is just like a linear equation ( $ax + b = 0$ ), but instead of an equals sign, you have one of these: $<$ , $>$ , $\le$ , or $\ge$ . The solution is not a single number — it's an interval (or a union of intervals) on the number line.

From Equation to Inequality

Start with a simple equation:

2x + 3 = 7

Solve it: $2x = 4 \implies x = 2$ . One point.

Now change it to an inequality:

2x + 3 > 7

Solve it the same way — but the meaning changes. Subtract 3: $2x > 4$ . Divide by 2: $x > 2$ .

The solution is all numbers greater than 2. On a number line, you draw an open circle at 2 (because 2 itself is not included) and shade everything to the right.

Watch out

If you multiply or divide both sides of an inequality by a negative number, the inequality sign reverses. For example: $-x < 5$ becomes $x > -5$ . This is the single most common mistake students make.

The Four Types of Solutions

Inequality	Meaning	Number line representation
$x > a$	All numbers strictly greater than $a$	Open circle at $a$ , shade right
$x \ge a$	All numbers greater than or equal to $a$	Closed (filled) circle at $a$ , shade right
$x < a$	All numbers strictly less than $a$	Open circle at $a$ , shade left
$x \le a$	All numbers less than or equal to $a$	Closed circle at $a$ , shade left

The solution set is usually written in interval notation:

$x > 2$ → $(2, \infty)$
$x \le -3$ → $(-\infty, -3]$

Parentheses $($ or $)$ mean the endpoint is not included. Brackets $[$ or $]$ mean it is included.

Solving a Linear Inequality: Step by Step

Solve $3x - 5 \le 7x + 3$ .

Bring variable terms to one side:

$3x - 5 - 7x \le 3$

$-4x - 5 \le 3$
Isolate the variable term:

$-4x \le 8$
Divide by the coefficient (here it's $-4$ , so reverse the sign):

$x \ge -2$

The solution is $x \ge -2$ , or in interval notation: $[-2, \infty)$ .

Tip

Always check your answer by testing a number from the solution set. For $x \ge -2$ , test $x = 0$ : $3(0)-5 = -5$ and $7(0)+3 = 3$ . Is $-5 \le 3$ ? Yes. Now test a number outside, say $x = -3$ : $3(-3)-5 = -14$ and $7(-3)+3 = -18$ . Is $-14 \le -18$ ? No — so the inequality fails, confirming our solution is correct.

Why This Matters …

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Expand both sides, collect like terms, and isolate $x$ to find that the inequality holds for all $x \le 4$ .

Linear inequalities behave almost exactly like equations when you solve them—with one critical exception. You can add, subtract, multiply, or divide both sides by any positive number without changing the direction of the inequality. The only time you flip the inequality sign is when you multiply or divide by a negative number.

Here we have a straightforward inequality with parentheses on both sides. The strategy is to expand, simplify, and isolate the variable.

Solution

Expand both sides using the distributive property.

Left side: $3(2 - x) = 6 - 3x$

Right side: $2(1 - x) = 2 - 2x$

So the inequality becomes:

6 - 3x \ge 2 - 2x

Collect all terms involving $x$ on one side.

Add $3x$ to both sides:

6 \ge 2 - 2x + 3x

6 \ge 2 + x

Isolate $x$ .

Subtract $2$ from both sides:

$4 \ge x$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 36 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$6(2x+3)+x>53-2x$
›Reveal solutionSolution

Expand and collect terms: $15x>35$ , so $x>\tfrac73$ .

Expand the left side:
$6(2x+3)+x=12x+18+x=13x+18.$
The inequality becomes …
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KEAM 2026Set eng-2026-04174 marksMCQ
$ABCD$
›Reveal solutionSolution

The perimeter is $18a+2$ ; solving $56<18a+2<92$ gives $3<a<5$ .

In the right-angled trapezium, $AB\parallel CD$ with the perpendicular leg $AD=3a$ . The horizontal offset between the parallel sides is $AB-CD=(7a+1)-(3a+1)=4a$ , so the slant side is
$BC=\sqrt{(4a)^2+(3a)^2}=\sqrt{25a^2}=5a.$
The perimeter is …
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KEAM 2026Set eng-2026-04184 marksMCQ
$x$
›Reveal solutionSolution

Solving $8+3x>4(x-3)+2$ yields $x<18$ .

$8+3x>4(x-3)+2=4x-12+2=4x-10.$
Bring terms together:

$8+10>4x-3x\;\Rightarrow\;18>x.$ …
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KEAM 2026Set eng-2026-04194 marksMCQ
$x$
›Reveal solutionSolution

Solve the two-sided inequality: $-14<x<-4$ or $6<x<16$ .

Upper bound $|x-1|<15$ : $-15<x-1<15\Rightarrow -14<x<16$ .

Lower bound $|x-1|>5$ : $x-1>5$ or $x-1<-5\Rightarrow x>6$ or $x<-4$ . …
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KEAM 2026Set eng-2026-04194 marksMCQ
$x$
›Reveal solutionSolution

Subtract 1 and analyse the sign: solution is $(-\infty,2)$ .

Rearrange.

$\frac{x-3}{x-2}-1\ge0\Rightarrow \frac{(x-3)-(x-2)}{x-2}\ge0\Rightarrow \frac{-1}{x-2}\ge0.$ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$\dfrac{2x - 10}{x + 2} \geq x - 5$
›Reveal solutionSolution

Factoring $2x-10=2(x-5)$ and moving everything to one side yields $-\frac{x(x-5)}{x+2}\ge0$ , i.e. $\frac{x(x-5)}{x+2}\le0$ . A sign chart gives $x<-2$ or $0\le x\le5$ .

Start from $\dfrac{2x-10}{x+2}\ge x-5$ . Since $2x-10=2(x-5)$ :
$\frac{2(x-5)}{x+2}-(x-5)\ge0\;\Rightarrow\;(x-5)\left[\frac{2}{x+2}-1\right]\ge0.$
Simplify the bracket:
$\frac{2-(x+2)}{x+2}=\frac{-x}{x+2},$
so
$(x-5)\cdot\frac{-x}{x+2}\ge0\;\Longleftrightarrow\;\frac{x(x-5)}{x+2}\le0.$
Critical points: $x=-2$ (excluded), $x=0$ , $x=5$ . Sign of $\frac{x(x-5)}{x+2}$ : …
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KEAM 2026Set eng-2026-04204 marksMCQ
$10 < |x + 10| \leq 25$
›Reveal solutionSolution

Substitute $y=x+10$ . The double inequality $10<|y|\le25$ means $y\in[-25,-10)\cup(10,25]$ . Converting back ( $x=y-10$ ) gives $x\in[-35,-20)\cup(0,15]$ .

Let $y=x+10$ , so the condition is $10<|y|\le25$ .

This breaks into:
$|y|>10 \Rightarrow y<-10 \text{ or } y>10,$
$|y|\le25 \Rightarrow -25\le y\le25.$ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$x$
›Reveal solutionSolution
Sign-analyse $\frac{(x-15)(x-70)}{x-30}\ge 0$ over integers $1$ – $100$ ( $x=30$ excluded).

Critical points: $15,30,70$ (with $x=30$ making the denominator zero).

$x<15$ : expression $<0$ .

$15\le x<30$ : $\ge 0$ (equals $0$ at $x=15$ ). Integers $15,\dots,29$ → $15$ values.

$30<x<70$ : $<0$ . …
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KEAM 2026Set eng-2026-04214 marksMCQ
$(x - 1)(x^2 - 5x + 7) < (x - 1)$
›Reveal solutionSolution

Bringing everything to one side factors as $(x-1)(x-2)(x-3)<0$ , giving $(-\infty,1)\cup(2,3)$ .

$(x-1)(x^2-5x+7)-(x-1)<0\;\Rightarrow\;(x-1)(x^2-5x+6)<0\;\Rightarrow\;(x-1)(x-2)(x-3)<0.$ …
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KEAM 2026Set eng-2026-04214 marksMCQ
$\left| x + \dfrac{1}{x} \right| > 2$
›Reveal solutionSolution

By AM–GM $|x+1/x|\ge2$ with equality only at $x=\pm1$ ; excluding those and $x=0$ gives $\mathbb{R}-\{-1,0,1\}$ .

For real $x\neq0$ , $\left|x+\dfrac1x\right|\ge2$ , with equality exactly when $x=\pm1$ . The strict inequality $>2$ therefore holds for all $x$ except $x=\pm1$ ; al …
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KEAM 2026Set eng-2026-04224 marksMCQ
$\frac{21x - 6}{4} - 9 \leq 0$
›Reveal solutionSolution

Solve each inequality separately and intersect.

First: $\frac{21x-6}{4}\le9\Rightarrow 21x-6\le36\Rightarrow 21x\le42\Rightarrow x\le2$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$|2x - 3| < 5, x \in \mathbb{R}$
›Reveal solutionSolution

Remove the modulus with a two-sided inequality; strict $<$ gives an open interval.

$|2x-3|<5$ means $-5<2x-3<5$ .

Add $3$ : $-2<2x<8$ . Divide by $2$ : $-1<x<4$ . …
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Q.3(2−x)≥2(1−x)3(2 - x) \ge 2(1 - x)3(2−x)≥2(1−x)

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Q. $3(2 - x) \ge 2(1 - x)$

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